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CAT 2018 Slot 2 QA Question & Solution

Number SystemsMedium

Question

If N and x are positive integers such that $N^{N}$ = $2^{160}\ and \ N{^2} + 2^{N}$ is an integral multiple of $\ 2^{x}$, then the largest possible x is

Solution

It is given that $N^{N}$ = $2^{160}$

We can rewrite the equation as $N^{N}$ = $(2^5)^{160/5}$ = $32^{32}$

$\Rightarrow$ N = 32

$N{^2} + 2^{N}$ = $32^2+2^{32}=2^{10}+2^{32}=2^{10}*(1+2^{22})$

Hence, we can say that $N{^2} + 2^{N}$ can be divided by $2^{10}$

Therefore, x$_{max}$ = 10